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GATE | GATE-CS-2017 (Set 1) | Question 65

  • Difficulty Level : Medium
  • Last Updated : 30 Sep, 2021

Instruction execution in a processor is divided into 5 stages. Instruction Fetch(IF), Instruction Decode (ID), Operand Fetch(OF), Execute(EX), and Write Back(WB), These stages take 5,4,20, 10 and 3 nanoseconds (ns) respectively. A pipelined implementation of the processor requires buffering between each pair of consecutive stages with a delay of 2 ns. Two pipelined implementations of the processor are contemplated:

(i) a naïve pipeline implementation (NP) with 5 stages and
(ii) an efficient pipeline (EP) where the OF stage id divided into stages OF1 and OF2 with execution times of 12 ns and 8 ns respectively.

The speedup (correct to two decimals places) achieved by EP over NP in executing 20 independent instructions with no hazards is ________________.

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Note: This questions appeared as Numerical Answer Type.
(A) 1.50-1.51
(B) 1.51-1.52
(C) 1.52-1.53
(D) 1.53-1.54


Answer: (A)

Explanation: Given, total number of instructions (n) = 20
For naive pipeline (NP):



Number of stages(k) = 5
Clock time (Tp) = max { (stage delay+buffer delay) } = { 7, 6, 22, 12, 5 } = 22 nsec
Execution time (Enp) = ( k + n - 1 )*Tp = ( 5 + 20 - 1 )*22 = 528 nsec


For efficient pipeline (EP):

number of stages(k) = 6 ( delay with 20 nsec stage is divided into 12 nsec and 8 nsec )
Clock time (Tp) = max { (stage delay+buffer delay) } = { 7, 6, 14, 10, 14, 5 } = 14 nsec
Execution time (Eep) =  ( k + n - 1 )*Tp = ( 6 + 20 - 1 )*14 = 350 nsec

Therefore,

Speedup = (Enp) / (Eep) = 528 / 350 = 1.508

This explanation is contributed by Mithlesh Upadhyay.




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